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## Multivariable calculus

### Course: Multivariable calculus>Unit 5

Lesson 8: Divergence theorem (articles)

# 3D divergence theorem examples

See how to use the 3d divergence theorem to make surface integral problems simpler.

## The divergence theorem (quick recap)

Setup:
• start color #0c7f99, start bold text, F, end bold text, end color #0c7f99, left parenthesis, x, comma, y, comma, z, right parenthesis is some three-dimensional vector field.
• start color #bc2612, V, end color #bc2612 is a three-dimensional volume (think of a blob in space).
• start color #bc2612, S, end color #bc2612 is the surface of start color #bc2612, V, end color #bc2612.
• start color #0d923f, start bold text, n, end bold text, with, hat, on top, end color #0d923f is a function which gives unit normal vectors on the surface of start color #bc2612, S, end color #bc2612.
Here's what the divergence theorem states:
$\displaystyle \underbrace{ \iiint_\redE{V} \text{div}\,\blueE{\textbf{F}}\,\redE{dV} }_{\substack{ \text{Add up little bits} \\\\ \text{of outward flow in \redE{V}} }} = \underbrace{ \overbrace{ \iint_\redE{S} \blueE{\textbf{F}} \cdot \greenE{\hat{\textbf{n}}} \,\redE{d\Sigma} }^{\text{Flux integral}} }_{\substack{ \text{Measures total outward } \\\\ \text{flow through \redE{V}'s boundary} }}$
The intuition here is that both integrals measure the rate at which a fluid flowing along the vector field start color #0c7f99, start bold text, F, end bold text, end color #0c7f99 is exiting the region start color #bc2612, V, end color #bc2612 (or entering start color #bc2612, V, end color #bc2612, if the values of both integrals are negative). Triply integrating divergence does this by counting up all the little bits of outward flow of the fluid inside start color #bc2612, V, end color #bc2612, while taking the flux integral measures this by checking how much is leaving/entering along the boundary of start color #bc2612, V, end color #bc2612.

## Strategizing

The divergence theorem lets you translate between surface integrals and triple integrals, but this is only useful if one of them is simpler than the other. In each of the following examples, take note of the fact that the volume of the relevant region is simpler to describe than the surface of that region.
In general, when you are faced with a surface integral over a closed surface, consider if it would be easier to integrate over the volume enclosed by that surface. If it is, it's a strong signal that the divergence theorem will come in handy.

## Example 1: Surface integral through a cube.

Problem
Let's say start color #bc2612, C, end color #bc2612 is a 1, times, 1, times, 1 cube, situated in space such that one corner is on the origin, one corner is at left parenthesis, 1, comma, 1, comma, 1, right parenthesis, and so that all its edges are parallel to one of the coordinate axes.
Let start color #bc2612, S, end color #bc2612 represents the surface of this cube, which consists of 6 square faces, oriented using outward facing normal vectors. Compute the following surface integral:
\iint, start subscript, start color #bc2612, S, end color #bc2612, end subscript, left parenthesis, 2, y, start bold text, i, end bold text, with, hat, on top, plus, 3, y, squared, start bold text, j, end bold text, with, hat, on top, plus, 4, z, start bold text, k, end bold text, with, hat, on top, right parenthesis, dot, start color #bc2612, d, \Sigma, end color #bc2612

Solution
Concept check: According to the divergence theorem, which of the following is equal to the surface integral we are asked to compute?

The cube is a great example of an object whose volume is simpler than its surface. To do this surface integral directly, you would have to address each of the 6 square faces separately. Furthermore, the vector-valued function we are integrating becomes simpler when we take the divergence, as you are about to see. So using the divergence theorem will be doubly helpful!
Concept check: Compute the divergence of the vector-valued function in the surface integral above.
del, dot, left parenthesis, 2, y, start bold text, i, end bold text, with, hat, on top, plus, 3, y, squared, start bold text, j, end bold text, with, hat, on top, plus, 4, z, start bold text, k, end bold text, with, hat, on top, right parenthesis, equals

Concept check: Use the divergence theorem to finish the problem by plugging the divergence you just computed to the triple integral you chose in the question before that:
integral, start subscript, 0, end subscript, start superscript, 1, end superscript, integral, start subscript, 0, end subscript, start superscript, 1, end superscript, integral, start subscript, 0, end subscript, start superscript, 1, end superscript, del, dot, left parenthesis, 2, y, start bold text, i, end bold text, with, hat, on top, plus, 3, y, squared, start bold text, j, end bold text, with, hat, on top, plus, 4, z, start bold text, k, end bold text, with, hat, on top, right parenthesis, d, x, d, y, d, z, equals

## Example 2: Surface integral through a cylinder

Problem
Let start color #bc2612, C, end color #bc2612 be a cylinder, whose base is a circle with radius 3 sitting on the x, y-plane centered at the origin, and whose height is 5.
Letting start color #bc2612, S, end color #bc2612 represent the surface of this cylinder, oriented with outward facing unit normal vectors compute the following surface integral:
$\displaystyle \iint_\redE{S} \left[ \begin{array}{c} x^3 \\ y^3 \\ x^3 + y^3 \end{array} \right] \cdot \redE{d\Sigma}$

Solution
As with the previous example, what signals that the divergence theorem might be useful is that the volume of our region is easier to describe than its surface. This is especially true if we anticipate integrating using cylindrical coordinates. And again, the divergence of the relevant function will make it simpler.
Concept check: Compute the divergence of the vector-valued function in the integral above.
$\displaystyle \nabla \cdot \left[ \begin{array}{c} x^3 \\ y^3 \\ x^3 + y^3 \end{array} \right] =$

Concept check: Compute the triple integral of this divergence inside the cylinder start color #bc2612, C, end color #bc2612 described in the problem. As a reminder, its base is a circle of radius 3 on the x, y-plane centered at the origin, and it has height 5.
$\displaystyle \iiint_{\redE{C}} \left( \nabla \cdot \left[ \begin{array}{c} x^3 \\ y^3 \\ x^3 + y^3 \end{array} \right] \right) \; \redE{dV} =$

## Example 3: Surface area from a volume integral

Problem:
Use the divergence theorem to compute the surface area of a sphere with radius 1, given the fact that the volume of that sphere is start fraction, 4, divided by, 3, end fraction, pi.

Solution
This feels a bit different from the previous two examples, doesn't it? To start, there is no vector field in the problem, even though the divergence theorem is all about vector fields!
If you let start color #bc2612, S, end color #bc2612 describe the surface of the sphere, its surface area will be given by the following as-simple-as-they-come surface integral:
\iint, start subscript, start color #bc2612, S, end color #bc2612, end subscript, 1, start color #bc2612, d, \Sigma, end color #bc2612
However, this is a surface integral of a scalar-valued function, namely the constant function f, left parenthesis, x, comma, y, comma, z, right parenthesis, equals, 1, but the divergence theorem applies to surface integrals of a vector field. In other words, the divergence theorem applies to surface integrals that look like this:
\iint, start subscript, start color #bc2612, S, end color #bc2612, end subscript, left parenthesis, start color #0c7f99, start bold text, F, end bold text, end color #0c7f99, dot, start color #0d923f, start bold text, n, end bold text, with, hat, on top, end color #0d923f, right parenthesis, start color #bc2612, d, \Sigma, end color #bc2612
Here, start color #0d923f, start bold text, n, end bold text, with, hat, on top, end color #0d923f is some function which gives unit normal vectors to start color #bc2612, S, end color #bc2612, the surface of the sphere. You can turn this into the desired surface area integral by finding a vector-valued function start color #0c7f99, start bold text, F, end bold text, end color #0c7f99 such that start color #0c7f99, start bold text, F, end bold text, end color #0c7f99, dot, start color #0d923f, start bold text, n, end bold text, with, hat, on top, end color #0d923f always equals 1.
Concept check: Which of the following definitions of a vector field start color #0c7f99, start bold text, F, end bold text, end color #0c7f99 will ensure the property start color #0c7f99, start bold text, F, end bold text, end color #0c7f99, dot, start color #0d923f, start bold text, n, end bold text, with, hat, on top, end color #0d923f, equals, 1 at all points on the surface of the sphere?

Concept check: Using this choice for start color #0c7f99, start bold text, F, end bold text, end color #0c7f99, together with the divergence theorem, which of the following integrals will give the surface area of the unit sphere? Let start color #bc2612, B, end color #bc2612 represent the volume enclosed by the sphere, also known as the "unit ball".

Concept check: Which of the following functions gives unit normal vectors to the surface of the unit sphere?